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Nina [5.8K]
3 years ago
5

3 by 4 rectangle is inscribed in a circle. what's the circumfrence

Mathematics
1 answer:
Darina [25.2K]3 years ago
7 0
The trick here is the recognize that the diagonal length of the rectangle is the diameter of the circle. 

W = width of rectangle = 4
L = length of rectangle = 3
D = diagonal of rectangle

Using pythagorean theorem, we solve for D.

W^2 + L^2 = D^2
(4)^2 + (3)^2 = D^2
16 + 9 = D^2
25 = D^2
5 = D or -5 = D

Since D is a length, D must be positive. Therefore, D=5.

D = diagonal of the rectangle = 5

Since D is also the diameter of the circle AND the diameter of a circle is twice the radius, we have the following equation :

r = radius of circle = 1/2 D = 1/2 (5) = 2.5
C = circumference

C = 2 pi r = 2 pi (2.5) = 5 pi
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cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0
Dafna11 [192]

Answer:

Radius=2.09 cm

Height,h=14.57 cm

Step-by-step explanation:

We are given that

Volume of cylinderical shaped can=200 cubic cm.

Cost of sides of can=0.02 cents per square cm

Cost of top and bottom of the can =0.07 cents per square cm

Curved surface area of cylinder=2\pi rh

Area of circular base=Area of circular top=\pi r^2

Total cost,C(r)=0.02\times 2\pi rh+2\pi r^2\times 0.07

Volume of cylinder,V=\pi r^2 h

200=\pi r^2 h

h=\frac{200}{\pi r^2}

Substitute the value of h

C(r)=0.02\times 2\pi r\times \frac{200}{\pi r^2}+2\pi r^2\times 0.07

C(r)=\frac{8}{r}+0.14\pi r^2

Differentiate w.r.t r

C'(r)=-\frac{8}{r^2}+0.28\pi r

C'(r)=0

-\frac{8}{r^2}+0.28\pi r=0

0.28\pi r=\frac{8}{r^2}

r^3=\frac{8}{0.28\pi}=9.095

r=(9.095)^{\frac{1}{3}}=2.09

Again, differentiate w.r.t r

C''(r)=\frac{16}{r^3}+0.28\pi

Substitute the value of r

C''(2.09)=\frac{16}{(2.09)^3}+0.28\pi=2.63>0

Therefore,the product cost is minimum at r=2.09

h=\frac{200}{\pi (2.09)^2}=14.57

Radius of can,r=2.09 cm

Height of cone,h=14.57 cm

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